Solving AB Beam Reactions with Hyperstatic System

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SUMMARY

The discussion focuses on solving the reactions of an AB beam subjected to a distributed load q over a length of 2 meters, using the differential equation y''=-(1/EI) *M(x). The participants identify six reactions (Va, Ha, Ma, Vb, Hb, Mb) at the supports and derive equations based on equilibrium conditions. The calculated reactions yield Va + Vb = 7.25 KN and a moment equation that requires further exploration. The problem is identified as statically indeterminate, necessitating additional methods such as deflection compatibility equations or virtual work for a complete solution.

PREREQUISITES
  • Understanding of beam theory and reactions in structural mechanics
  • Familiarity with differential equations, specifically y''=-(1/EI) *M(x)
  • Knowledge of equilibrium equations in statics
  • Concepts of statically indeterminate structures
NEXT STEPS
  • Study methods for solving statically indeterminate beams, including virtual work
  • Learn about deflection compatibility equations in structural analysis
  • Explore the application of the moment-curvature relationship in beam theory
  • Investigate numerical methods for solving differential equations related to beam reactions
USEFUL FOR

This discussion is beneficial for civil engineers, structural analysts, and students studying mechanics of materials, particularly those dealing with beam reactions and statically indeterminate systems.

Zouatine
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hello everyone
1. Homework Statement

Problem:
An AB beam, loaded with a distributed load q (KN/m), the length of the beam is 2 m.
EI=constant
p_109797omn1.png

find the reactions in the beam ,we use y''=-(1/EI) *M(x)

Homework Equations


y''=-(1/EI) *M(x)[/B]

The Attempt at a Solution


first we have 6 reactions (Va,Ha,Ma,Vb,Hb,Mb) in the supports.
∑F/x=0→ Ha+Hb=0 , Ha=-Hb=0 -------------1
∑F/y=0→ Va+Vb-Q=0 → Va+Vb=Q→ Va+Vb=∫ ((e^x)+(e^-x))dx 0≤ x ≤2
Va+Vb= 7,25 KN -------------2
∑M/b=0 → 2*Va- Q*(center of gravity of the load q)-Ma+Mb=0
center of gravity of the load q = (∫∫x*ds)/(∫∫ds)
∫∫x*ds= ∫∫x*dx*dy 0≤y≤(e^x)+(e^-x) , 0≤x≤2
∫∫x*ds= ∫∫x*dx*dy = 9,25 KN*m
∫∫ds= 7,25 KN
center of gravity of the load q = (∫∫x*ds)/(∫∫ds)=1,28 m (2-1,28)= 0,72m
2*Va- Q*0,72-Ma+Mb=0
2*Va-Ma+Mb=5,22 KN m----------3
Differential equation:
y''=-(1/EI) *M(x)
M(x) = ?? 0≤x<2
my problem is how to find the moment equation?
thanx
 

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Zouatine said:
hello everyone
1. Homework Statement

Problem:
An AB beam, loaded with a distributed load q (KN/m), the length of the beam is 2 m.
EI=constant
View attachment 236680
find the reactions in the beam ,we use y''=-(1/EI) *M(x)
2. Homework Equations
y''=-(1/EI) *M(x)
3. The Attempt at a Solution

first we have 6 reactions (Va,Ha,Ma,Vb,Hb,Mb) in the supports.
∑F/x=0→ Ha+Hb=0 , Ha=-Hb=0 -------------1
∑F/y=0→ Va+Vb-Q=0 → Va+Vb=Q→ Va+Vb=∫ ((e^x)+(e^-x))dx 0≤ x ≤2
Va+Vb= 7,25 KN -------------2
∑M/b=0 → 2*Va- Q*(center of gravity of the load q)-Ma+Mb=0
center of gravity of the load q = (∫∫x*ds)/(∫∫ds)
∫∫x*ds= ∫∫x*dx*dy 0≤y≤(e^x)+(e^-x) , 0≤x≤2
∫∫x*ds= ∫∫x*dx*dy = 9,25 KN*m
∫∫ds= 7,25 KN
center of gravity of the load q = (∫∫x*ds)/(∫∫ds)=1,28 m (2-1,28)= 0,72m
2*Va- Q*0,72-Ma+Mb=0
2*Va-Ma+Mb=5,22 KN m----------3
Differential equation:
y''=-(1/EI) *M(x)
M(x) = ?? 0≤x<2
my problem is how to find the moment equation?
thanx
i haven’t checked your calculus, but you seem to be on the right track. However, are you sure that both ends are fixed? If so, the problem is statically indeterminate to the second degree, and you have to resort to other equations besides the equilibrium equations, like deflection compatibility equations, virtual work, or other methods, which is a bit tedious.
 

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